Image of Homomorphisms for Cyclic Groups

I'm reading an example for abstract algebra and I would like to clarify some terms. The example is showing the number of homomorphisms from $\mathbb Z$ to $S_3$.

Take a homomorphism $f$ from $\mathbb Z$ to $S_3$. Since $1$ is a generator of $\mathbb Z$, $f$ is completely determined by the image of $1$. Since $1$ has infinite order, there are no restrictions on the image of $1$. Thus each element of $S_3$ is a possible image for $1$. There are $6$ homomorphisms from $\mathbb Z$ to $S_3$.

What does "completely determined by the image of $1$" mean? I thought about it a little bit, and it seems that since $1$ is a generator of $\mathbb Z$, any element in $\mathbb Z$ can be generated from $1^k$. When it says image of $1$, does it mean $f(1^k)$, where $k$ is an integer? In general, what does the "image" of an element under homomorphism mean? I thought for an element $g$ in group $G$, and for a homomorphism $h$, the image of $g$ under $h$ is $h(g)$. How can that determine the entire homomorphism?

Thanks for your help.

• As $\mathbb Z$ is written additively, it qwould be less confusing to write $k\cdot 1$ instead of $1^k$ (after all, it is supposed to mean $\underbrace{1+1+\ldots+1}_{k}$ – Hagen von Eitzen Oct 19 '14 at 20:32
• Thanks Hagen. How did you fix up the formatting? Would like to learn for future posts. – jstnchng Oct 19 '14 at 20:52